Question 1: Two angles of a triangle measure 58^{\circ} \ and\  87^{\circ} . Find the measure of the third angle.

Answer:

Sum of all the angles of a triangle = 180^{\circ} 

Let the third angle = x 

Therefore 58 + 87 + x = 180^o

Or x= 35^{\circ} 

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Question 2: Three angles of a triangle are in the ratio of 2 : 5 : 8 .  Find the angles of the triangle.

Answer:

Let the angles be 2x, 5x, \ and\  8x 

Therefore 2x+5x+8x = 180^o 

Or x= 12^o 

Therefore the angles are 24^{\circ}, 60^{\circ}, \ and\  96^{\circ}. 

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Question 3: One angle of the triangle is 76^o  and the other two angles are equal. Find the measure of each angle of the triangle.

Answer:

Let the two equal angles be x^{\circ} 

Therefore 76 + 2x = 180^o 

Or x = 52^o 

Therefore the angles are 76^{\circ}, 52^{\circ} \ and\  52^{\circ}. 

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Question 4: One angle of the triangle is 85^{\circ}   and the other two angles are in the ratio 8 :11 Find the measures of each of the unknown angles.

Answer:

Let the two unknown angles be 8x \ and\  11 x 

Therefore 85+8x+11x = 180^o 

Or 19x = 95 

Or x = 5^o 

Therefore the two angles are 40^{\circ} \ and\  55^{\circ}. 

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Question 5: Three angles of a triangle measures (2x)^{\circ}, (4x-7)^{\circ} \ and\  (5x-11)^{\circ} .  Find the value of x   and the measures of all the angles.

Answer:

2x + (4x-7) + (5x-11) = 180^o 

Or x = 18^o 

Hence the angles are 36^{\circ}, 65^{\circ}, \ and\  79^{\circ} 

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Question 6: In  \Delta ABC, \angle A=x^{\circ}, \angle B=(2x-15)^{\circ}, \ and\  \angle C=(3x+21)^{\circ} . Find the value of x and the measure of all the angles of the triangle.

Answer:

x+(2x-15)+(3x+21)=180^o 

or x=29^o 

Hence \angle A=29^{\circ}, \angle B=43^{\circ}, \ and\  \angle C=108^{\circ} 

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Question 7: In \Delta ABC, \angle A=(2x+15)^{\circ}, \angle B=x^{\circ}, \ and\  \angle C=(3x-15)^{\circ} . Find the value of x and the measure of all the angles of the triangle and show that the triangle is an isosceles triangle.

Answer:

(2x+15)+x+(3x-15)=180^o \ or\  x=30^o 

Hence \angle A=75^{\circ}, \angle B=30^{\circ}, \ and\  \angle C=75^{\circ}.  

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Question 8: In \Delta ABC, \angle A=3\ \angle B=6 \ \angle C  . Find the angles of the triangle.

Answer:

Let \angle A=3\angle B=6 \angle C=6x 

\Rightarrow  \angle A=6x,  \angle B=2x, and \angle C=x 

\Rightarrow  6x+2x+x=180^o \ or\  x=20^o 

\Rightarrow   \angle A=120^{\circ}, \angle B=40^{\circ} \ and\  \angle C=20^{\circ}   

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Question 9: In the adjoining figure, it is given that \angle ABC=40^{\circ}, \angle ACD=70^{\circ}, \angle ACB=x^{\circ} \ and\  \angle EAC=y . Find the value of $latex x and y .

Answer:p10

x+70=180^o \ hence\  x=110^{\circ} 

Note: Sum of angles of adjacent angles on a straight line =180^o 

\Rightarrow 40+110+(180-y)=180^o 

Note:Sum of all angles of a triangle =180^o 

\Rightarrow y=150^{\circ} 

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Question 10: In the adjoining figure, it is given that \angle A=50^{\circ} \ and\  \angle B=70^{\circ} and BO \ and\  CO  are bisectors of \angle B \ and\  \angle C respectively. Find the measure of \angle BOC.

Answer:P11

Let \angle OCA= \angle OCB=x^{\circ}

Since BO \ bisects \angle A=ABC, \angle ABO= \angle OBC=35^{\circ}.

\Rightarrow 50^o+70^o+2x=180^o

\Rightarrow  x=30^{\circ}=\angle OCB

\Rightarrow 35^o+30^o+ \angle BOC=180^o

\Rightarrow  \angle BOC=115^{\circ}

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Question 11: Using the information given in the adjoining figure, find the value of x \ and\  y. P12

Answer:

36^o+y+x=180^o

(24^o+36^o)+2x+64^o=180^o

\Rightarrow  x=28^{\circ}

\Rightarrow 36^o+y+28^o=180^o \ or\  y=116^{\circ}

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Question 12: In the adjoining figure, DCBE is a straight line, \angle ABD=135^{\circ} \ and\  \angle ACE=125^{\circ} . Find \angle BAC.

Answer:P13

 \angle ABC=180^o-135^o=45^{\circ}

\angle ACB=180^o-125^o=55^{\circ}

Therefore \angle BAC=180^o- \angle ABC- \angle ACB

=180^o-45^o-55^o=80^o

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Question 13: Find, giving reasons, the unknown angles marked by x, \ y in the following figures.

Answer:P14

i) 58^o+x+(180^o-123^o)=180^o

\Rightarrow x=65^{\circ}

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ii) 42^o+y=180^o

\Rightarrow y=138^{\circ}

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iii) x+72^o+42^o=180^o

\Rightarrow x=68^{\circ}

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iv) 50^o+x+x=180^o

\Rightarrow x=65^{\circ}

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v) 65^o+y=180^o

\Rightarrow y=115^{\circ}

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vi) 30^o+y+110^o=180^o

\Rightarrow y=40^{\circ}

Given AC=AD which means \angle ACD= \angle ADC=70^{\circ}

x+70^o+70^o=180^o

\Rightarrow x=40^{\circ}

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Question 14: In the adjoining figure, if \angle A=(x+5)^{\circ}, \angle B=(2x+3)^{\circ} \ and\  \angle BCD=(5x-14)^{\circ} , find the value of x and hence, find the measure of   i)\  \angle BCD \ ii)\  \angle ACB

Answer:P15

\angle A+ \angle B+ \angle C=180^{\circ}

\Rightarrow (x+5)+(2x+3)+{180^o-(5x-14)}=180^o

\Rightarrow x=11 \ Therefore\  \angle A=16^{\circ}, \angle B=25^{\circ}

\Rightarrow  i) \angle BCD=41^{\circ} \ \ and \ \ ii) \angle ACB=180^o-41^o=139^{\circ}

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Question 15: In \Delta ABC , side AC has been produced to D. If \angle BCD=125^{\circ} \ and\  \angle A : \angle B=2 :3 , find the measure of \angle A \ and\  \angle B.

Answer:P16

Let \angle A=2x \ and\  \angle B=3x.

\Rightarrow 2x+3x+(180^o-125^o)=180^o

\Rightarrow x=25^o

Therefore \angle A=50^{\circ} \ and\  \angle B=75^{\circ}

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Question 16: In the adjoining figure, it is given that: BC \parallel DE,\ \angle BAC=35^o \ and\  \angle BCE=102^o . Find the measure of i) \angle ABC ii)\  \angle ADE \ and\  iii) \angle CED

Answer:P17

i) \angle ABC+35^o+(180^o-102^o)=180^o

\Rightarrow  \angle ABC=67^{\circ}

\angle DEA+102^o=180^o   Note:Co-interior Angles

\Rightarrow  \angle DEA=78^{\circ}

ii) \angle ADE+78^o+35^o=180^o

\Rightarrow  \angle ADE=67^{\circ}

iii) \angle CED+67^o+35^o=180^o

\Rightarrow  \angle CED=78^{\circ}

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Question 17: In the adjoining figure, it is given that:AB=AC, \angle BAC=36^{\circ}, \angle ADB=45^{\circ} and \angle AEC=40^{\circ} . Find i) \angle ABC, ii) \angle ACB iii) \angle DAB and iv) \angle EAC

Answer:P18

AB=AC \Rightarrow \angle ABC= \angle ACB=x

Therefore 36^o+2x=180^o \Rightarrow x=72^o

i) & ii) Hence \angle ABC= \angle ACB=72^{\circ}

iv) \angle CAE+(180^o-72^o)+40^o=180^o

\Rightarrow \angle CAE=32^{\circ}

iii) \angle DAB+(180^o-72^o)+4^o5=180^o

\Rightarrow \angle DAB=27^{\circ}

\angle DAE=27^o+36^o+32^o=95^{\circ}

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Question 18: Using the information given in the adjoining figure, calculate the values of x \ and\  y. P19

Answer:

In \Delta ABC

34+(180^o-x)+62^o=180^o

\Rightarrow x=96^{\circ}

In \Delta DCE 24^o+96^o+(180^o-y)=180^o

\Rightarrow y=120^{\circ}

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Question 19: Sides AB \ and\  AC of \Delta ABC  have been produced to D and E respectively. BI and CI are bisectors of \angle CBD and \angle BCE respectively. If  \angle BAC=70^{\circ} and \angle ACB 50^{\circ} , find the measure of \angle BIC P20

Answer:

Given \angle DBI= \angle IBC=x and \angle BCI= \angle ICE=y

\therefore 50^o+2y=180^o \Rightarrow y=65^o

\angle ABC+50^o+70^o=180^o

\Rightarrow \angle ABC=60^{\circ}

60^o+2x=180^o

\Rightarrow x=60^o

\therefore 60^o+ \angle BIC+65^o=180^o

\Rightarrow \angle BIC=55^{\circ}     

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Question 20: In the adjoining figure, it is being given that BC \parallel DE,  \angle CED=70^{\circ}, \angle CBA=84^{\circ} \ and\  \angle BAC=x^{\circ} .  Find the value of x .

Answer:P21

\angle DEC= \angle BCA=70^{\circ} Note:Alternate angles

x+84^o+70^o=180^o \Rightarrow x=26^{\circ}

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Question 21: In the adjoining figure, it is being given that BC \parallel DE, \angle EDC=25^{\circ}, \angle ECD=20^{\circ}, \angle ABC=70^{\circ}  \angle BAC=x^{\circ} and \angle DEA=y^{\circ} . Find the value of x \ and\  y .

Answer:P22

In \Delta DEC,

25^o+20^o+ \angle DEC=180^o

Rightarrow  \angle DEC=135^{\circ}

\therefore y+135^o=180^o

\Rightarrow y=45^{\circ}

 BC \parallel DE \Rightarrow   \angle CBD= \angle EDA=70^{\circ}

In \Delta ADE,

70^o+45^o+x=180^o

\Rightarrow x=65^{\circ}

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Question 22: In the adjoining figure, it is given that \angle ACB=42^{\circ}, \angle DEB= \angle CAB=90^{\circ}, \angle ABC=x^{\circ},  \\ \angle BDE=y^{\circ} \ and\  \angle AFE=z^{\circ}. Find the value of x, \ y \ and \  z.

Answer:

In P23\triangle FEC , 90^o+42^o+(180^o-z) = 180^o

\Rightarrow z=132^{\circ}

In \Delta ADF, (180^o-132^o)+90^o+y=180^o

\Rightarrow y=42^{\circ}

In \Delta DBE, 42^o+90^o+x=180^o

\Rightarrow x=48^{\circ} 

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Question 23: In the adjoining figure, AE \parallel BC, \angle DAE=x^{\circ}, \angle ACB=(x-15)^{\circ}, \\ \angle BAC=(\frac{x}{2}+y)^{\circ} \ and\  \angle ABC=(y+15)^{\circ}

Answer:P24

In \Delta ABC, (y+15)+(x-15)+(\frac{x}{2}+y)=180^o

\Rightarrow 2y+\frac{3x}{2}=180^o ... ... ... ...(i)

Since AE \parallel BC,

x=(y+15) ... ... ... ...ii) Note:Alternate Angles

Substituting ii) in i) we get

2y+\frac{3}{2} (y+15)=180^o

\Rightarrow 4y+3y+45=360^o \Rightarrow y=45^{\circ}

Substituting the value of y \ in \ i) x=45+15=60^{\circ} 

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Question 24: In the adjoining figure, BA \parallel CD . Find the values of x \ and\  y. P25

Answer:

In \Delta ABC

(y+7)+(y-10)+(4x-5)=180^o

\Rightarrow 2y+4x=188^o ... ... ... ... (i)

Since BA \parallel CD,

(y-10)=5x ... ... ... ...(ii)

Substituting ii) in i) we get

2(5x+10)+4x=188^o

\Rightarrow 14x=168^o

\Rightarrow x=   12^{\circ} 

Substituting in ii) we get     y=70^{\circ}

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